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GuDViN [60]
4 years ago
15

Jerry measured the mass of the rubber ball to be 12.5 g. His calculated volume was 10.3 cm3. Using the formula density = mass/vo

lume, and the correct number of significant digits, Jerry should record the density as........ 1.21 g/cm3 1.2 g/cm3 1 g/cm3 1.214 g/cm3
Physics
1 answer:
solmaris [256]4 years ago
5 0
The answer is 1.21 grams per cm cubed
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Answer:

x = -6.5 meters

Explanation:

The position of a ball as a function of time t is given by :

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We need to find the position of the ball at 1.9 s. It can be simply calculated putting t = 1.9 s in equation (1) as :

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x = -6.5 meters

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What is the wavelength, in nm, of a photon with energy
lara31 [8.8K]

Answer:

(a)  λ = 4136 nm → infrared

(b) λ = 413.6 nm → visible light

(c) λ = 41.36 nm → ultraviolet

Explanation:

The wavelength of infrared is on the range of 700 nm to 1000000 nm

The wavelength of visible light is between 400 nm and 700 nm

The wavelength of ultraviolet ray on the range of 10 nm to 400 nm

The wavelength of photon is given by;

E = hf

f is the frequency of the wave = c / λ

E = h\frac{c}{\lambda}\\\\ \lambda = \frac{hc}{E}

Where;

c is the speed of light = 3 x 10⁸ m/s

h is Planck's constant = 6.626 x 10⁻³⁴ J/s

(a) 0.3 eV = 0.3 x 1.602 x 10⁻¹⁹ J

\lambda = \frac{(6.626 * 10^{-34})(3*10^8)}{(0.3)*(1.602*10^{-19})}\\\\\lambda = 4.136 *10^{-6} \ m

λ = 4136 x 10⁻⁹ m

λ = 4136 nm → infrared

(b) 3.0 eV

\lambda = \frac{(6.626 * 10^{-34})(3*10^8)}{(3)*(1.602*10^{-19})}\\\\\lambda = 4.136 *10^{-7} \ m

λ = 413.6 x 10⁻⁹ m

λ = 413.6 nm →visible light

(c) 30 eV

\lambda = \frac{(6.626 * 10^{-34})(3*10^8)}{(30)*(1.602*10^{-19})}\\\\\lambda = 4.136 *10^{-8} \ m

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